question_answer

If m is chosen in the quadratic equation \[({{m}^{2}}+1){{x}^{2}}-3x+{{({{m}^{2}}+1)}^{2}}=0\]such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :-             [JEE Main 9-4-2019 Afternoon]

A) \[8\sqrt{3}\]                            

B) \[4\sqrt{3}\]

C) \[10\sqrt{5}\]              

D) \[8\sqrt{5}\]

Correct Answer: D

Solution :

\[SOR=\frac{3}{{{m}^{2}}+1}\Rightarrow {{(S.O.R)}_{\max }}=3\]           when\[m=0\] \[\alpha +\beta =3\] \[\alpha \beta =1\] \[|{{\alpha }^{3}}-{{\beta }^{2}}|=||\alpha -\beta |({{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta )|\] \[\left| \sqrt{\alpha -\beta {{)}^{2}}-\alpha \beta }({{(\alpha +\beta )}^{2}}-\alpha \beta ) \right|\] \[=\left| \sqrt{9-4}(9-1) \right|\] \[=\sqrt{5}\times 8\]